And 3. The energy of the deuterium is not calculated relativistic but the γ-ray is completely relativistic, we then get: [math]m_{d}v_{d}=\frac{E_{\gamma}}{c}\rightarrow m_{d}^{2}=\frac{E_{\gamma}^{2}}{c^{2}}\rightarrow \frac{1}{2}m_{d}^{2}v_{d}^{2}=\frac{E^{2}_{\gamma}}{2c^{2}}[/math]

Inserting the energy of the deuterium we get:

[math]E_{d}=\frac{E_{\gamma}^{2}}{2m_{d}c^{2}}[/math]

we know that [math]E_{d}+E_{\gamma}=Q=2.2224\, MeV[/math] solving for this:

[math]E_{d}=0.0013\, MeV,\, E_{\gamma}=2.2226\, MeV[/math]

3: Q-Values

40Ca: M(40Ca)+M(α)-M(44Ti)=5.126 MeV

22Cr: 7.61 MeV

56Fe: 6.29 MeV

58Ni: 3.367 MeV

4: For ground state transitions assuming zero momentum to the neutrino

0.16 MeV

0.78 MeV

1.655 MeV subtracting the mass of a electron and a positron we end up at 0.633 MeV.

β- :0.579 for 64Zn, β+ : 0.653 MeV for 64Ni.

The cause of this effect is the even-even/odd-odd addition to the energy

228Ra is created with desintigration of 232Th: 232Th arrow 228ra + α +Q this givesM(228Ra) = M(230Th)-M(α)-QAll of the naural thorium that exist is 232Th. The mass that is given for thorium is therefore more or less equal to the mass 232Th 232.0381 u. It is much the same for helium where the mass is 4.002602 u. To find M(228Ra) we need to find the Q-value. The excess energy (Q-value) is distributed as kinetic energy on the products 228Ra and alpha. The alpha particle has very little mass compared to 228Ra, and the alpha particle by conversion of momentum will receive almost all the kinetic energy: